Local gravity calculation

One term I fully understand yet I have never seen how one actually does the calculation is the local gravity a particle feels in a gravitational field.

Now, I honestly feel this is as stupid of a question as they come, but intuitively I'd say, if I wanted a(r), the acceleration as a function of the radius (useful for things like Hawking temperature), you simply take your geodesic equation

The 'local gravity' is the coordinate acceleration of a freely falling particle, in your example d2xi/dλ2. And in turn 'coordinate acceleration' means with respect to a distinguished system of coordinates, such as the stationary coordinates in a stationary solution.

to find the acceleration as a function of the radial distance, multiply by the redshift, and do your usual simple radially-infalling particle and wala, local acceleration.

Sorry to be pedantic, but my wife is a French teacher, and I want to make sure she doesn't choke on a cherry pit and die if she sees this. It's "voilà," not "wala." Some French speakers sometimes pronounce it in certain speciific contexts without the "v" sound, but it's not normal: http://forum.wordreference.com/showthread.php?t=465332 It would be sort of like an American saying "I'm a," instead of "I'm gonna."

Sorry to be pedantic, but my wife is a French teacher, and I want to make sure she doesn't choke on a cherry pit and die if she sees this. It's "voilà," not "wala." Some French speakers sometimes pronounce it in certain speciific contexts without the "v" sound, but it's not normal: http://forum.wordreference.com/showthread.php?t=465332 It would be sort of like an American saying "I'm a," instead of "I'm gonna."

Well apparently there was all sorts of wrong going on in my thread so might as well butcher some languages as well.

I'll try to describe, briefly, the super-simple method for frame fields. In a rather non-rigorous way.

Let's start with polar coordinates, in flat space time, which should be familiar. You have some metric

ds^2 = -dt^2 + r^2 + r^ d theta^2

Because of the coefficient of r^2, [itex]d\theta[/itex] doesn't represent a constant distance.

What you do is you introduce locally, some vectors [itex]\hat{r}[/itex] = dr, and [itex]\hat{\theta} = r d\theta[/itex] that are unit length. At this point, I'm not worring about whether the "vector" dr is covariant or contravariant, though it turns out to be the former, when you think of it as a vector at all, that is.

More formal treatments will distinguish the coframes , dr, from the frames [itex]\partial / \partial r[/itex], I'm beeing very lax by glossing over this. It's eventually important to understand the fine distinctions here, but it's not good if it distracts you from understanding the basic idea of what's going on. It's the whole covariant/ contravariant mess...

You are essentially introducing new coordinates nearby any local event, the new coordinates have hats.

You can think of it as doing a coordinate transformation. There are lots of ways you coulc do a coordinate transformation, for instance going back to the definition of the metric as a tensor, but the easiest one is algebra:

You visualize [itex]\hat{r}, \hat{\theta}[/itex] as unit vectors, little unit arrows, that form a local coordinate system .

And you just do the same thing in Schwarzschild coordinates. You've got non-unity metric coefficeints

gtt dt^2 + grr dr^2

so you just define local "vectors" [itex]\hat{t}[/itex] = sqrt(|g11|) dt and [itex]\hat{r}[/itex] = sqrt(|grr|) dr

and you proceed on just as you did in polar coordinates, and you interpret the symbols the same way, the little hats are local vectors, that form their own little local coordinate system, which is orthonormal and very familiar and easy to deal with.